A zinc rod is dipped in 0.1 M solution of `ZnSO_(4)`. The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential `(E_(Zn^(2+)//Zn)^(@)=-0.76V)`

A zinc rod is dipped in 0.1 M solution of `ZnSO_(4)`. The salt is 95%

1.	A zinc rod is dipped in 0.1 M solution of `ZnSO_(4)`. The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential `(E_(Zn^(2+)//Zn)^(@)=-0.76V)`
Answer» The electrode reaction written as reduction reaction is: `Zn^(2+)+2e^(-)Zn(n=2)` Applying nernst equation, we get `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.0591)/(2)"log"(1)/([Zn^(2+)])` As 0.1 M `ZnSO_(4)` solution is 95% dissociated, this means that in the solution, `[Zn^(2+)]=(95)/(100)xx0.01M=0.095M` `thereforeE_(Zn^(2+)//Zn)=-0.76-(0.0591)/(2)"log"(1)/(0.095)=0.76-0.02955(log1000-log95)` `=0.76-0.02955(3-1.9777)=-0.76-0.03021=-0.79021` volt

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A zinc rod is dipped in 0.1 M solution of `ZnSO_(4)`. The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential `(E_(Zn^(2+)//Zn)^(@)=-0.76V)`

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